The circuit given resembles the balanced Wheatstone Bridge as $\frac{4}{6}=\frac{2}{3}$.
Thus, middle arm containing $4 \Omega$ resistance will be ineffective and no current flows through it. The equivalent circuit is shown as below :
Net resistance of $A B$ and $B C$
$R'=4+2=6 \Omega$
Net resistance of $A D$ and $DC$
$R''=6+3=9 \Omega$
Thus, parallel combination of $R'$ and $R$ " gives
$R=\frac{R' \times R''}{R'+R''} $
$=\frac{6 \times 9}{6+9}=\frac{54}{15}=\frac{18}{5} \Omega$
Hence, current $i=\frac{V}{R}=\frac{V}{18 / 5}=\frac{5 V}{18}$
